Solve for $x$ and $y$ using elimination. $\begin{align*}-8x+6y &= 6 \\ -4x+8y &= -2\end{align*}$
Answer: We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-1$ and the bottom equation by $2$ $\begin{align*}8x-6y &= -6\\ -8x+16y &= -4\end{align*}$ Add the top and bottom equations. $10y = -10$ Divide both sides by $10$ and reduce as necessary. $y = -1$ Substitute $-1$ for $y$ in the top equation. $-8x+6( -1) = 6$ $-8x-6 = 6$ $-8x = 12$ $x = -\dfrac{3}{2}$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = -1$.